Which of the following functions is differentiable at `x = 0?`

Options

A. `cos (|x|)+|x|`
B. `cos (|x|)-|x|`
C. `sin (|x|)+|x|`
D. `sin (|x|)-|x|`

Answer

D

Explanation

We have
`cos |x|={{:(,cos x,x ge 0),(,cos (-x)=cos x,x lt 0):}`
`Rightarrow cos |x|=cosx " for all x"in R`
Similarly, we have
`sin |x|={{:(,sin x,x ge0),(,sin(-x)=-sin x,x lt0):}`
Let `f(x)=cos(x)+|x|and g(x)=cos |x|-|x|`
Since cos x is everywhere differentiable and |x| is not differentiable at x=0. Therefore, f(x) and g(x) are not differentiable at x=0.
`Let u=sin(|x|)+|x|`, Then, `u(x)={{:(,sin x+x,x ge 0),(,-sin x-x,x lt 0):}`
`therefore ("LHD at x=0")={(d)/(dx)(-sin x-x)}_(at x=0)`
`Rightarrow ("LHD at x=0")-(-cos x-1)_(at x=0)=-cos 0-1=-2` and
`("RHD at x=0")={(d)/(dx)(sin x+x)}_(at x=0)=(cos x+1)_(at x=0)=2`
Clearly, (LHD atx=0)`ne ("RHD at x=0")`
So, u(x) is not differntiable at x=0
`"Let "v(x)=sin |x|-|x|."Then"`
`v(x)={{:(,-sinx+x,x lt 0),(,sin x-x,x ge0):}`
`therefore ("LHD at x=0")={(d)/(dx) (-sin x+x)}_("at x=0")`
`Rightarrow ("LHD at x=0")=(-cosx+1)_(at x=0)=-1+1=0`
`therefore ("RHD at x=0")=(cos x-1)_(at x=0)=cos 0-1=0`
`therefore ("LHD at x=0")=("RHD at x=0")`
So, v(x) is differentiable at x=0.